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After successive failures in the battles against the Union, the Empire retreated to its last stronghold. Depending on its powerful defense system, the Empire repelled the six waves of Union‘s attack. After several sleepless nights of thinking, Arthur, General of the Union, noticed that the only weakness of the defense system was its energy supply. The system was charged by N

nuclear power stations and breaking down any of them would disable the system.

The general soon started a raid to the stations by N special agents who were paradroped into the stronghold. Unfortunately they failed to land at the expected positions due to the attack by the Empire Air Force. As an experienced general, Arthur soon realized that he needed to rearrange the plan. The first thing he wants to know now is that which agent is the nearest to any power station. Could you, the chief officer, help the general to calculate the minimum distance between an agent and a station?

Input

The first line is a integer T representing the number of test cases.
Each test case begins with an integer N (1 ≤ N ≤ 100000).
The next N lines describe the positions of the stations. Each line consists of two integers X (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the station.
The next following N

lines describe the positions of the agents. Each line consists of two integers X (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the agent. 　

Output

For each test case output the minimum distance with precision of three decimal placed in a separate line.

Sample Input

2
4
0 0
0 1
1 0
1 1
2 2
2 3
3 2
3 3
4
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0

Sample Output

1.414
0.000


題目分析 ： 給你一堆點，找出不在同一集合中最近的兩個點

思路分析 ： 經典的分治板子題，註意剛剛學到的函數 dis 的寫法，可用性得到了提高

代碼示例 ：

#define ll long long
const ll maxn = 2e5+5;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;
ll n;

struct node
{
    double x, y;
    ll pt;
}pre[maxn], f[maxn];

bool cmpx(node a, node b){
    if (a.x == b.x) return a.y < b.y;
    else return a.x < b.x;
}

bool cmpy(node a, node b){
    return a.y < b.y;
}

double dis(node a, node b){ 
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double abc(double x){
    return x < 0?-x:x;
}

double close_pair(ll l, ll r){
    double d = 1.0*inf;
    if (l == r) return d;
    if (l + 1 == r){
        if (pre[l].pt != pre[r].pt) return dis(pre[l], pre[r]);
        else return d;
    }
    ll m = (l + r) >> 1;
    double d1 = close_pair(l, m);
    double d2 = close_pair(m+1, r);
    d = min(d1, d2);
     
    ll k = 1;
    for(ll i = l; i <= r; i++){
        if (abc(pre[m].x-pre[i].x) <= d) {f[k].x = pre[i].x; f[k].y = pre[i].y; f[k++].pt = pre[i].pt;} 
    }
    sort(f+1, f+k, cmpy);
    for(ll i = 1; i < k; i++){
        for(ll j = i+1; j < k && (f[j].y-f[i].y <= d); j++){
            if (f[i].pt == f[j].pt) continue;
            double ddd = dis(f[i], f[j]);
            d = min(d, ddd);
        }
    } 
    return d;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll t;
    double x, y;
    
    cin >> t;
    while(t--){
        cin >> n;
        for(ll i = 1; i <= n; i++){
            scanf("%lf%lf", &x, &y);
            pre[i].x = x; pre[i].y = y; pre[i].pt = 1;
        }    
        for(ll i = n+1; i <= 2*n; i++){
            scanf("%lf%lf", &x, &y);
            pre[i].x = x; pre[i].y = y; pre[i].pt = 2;
        }
        sort(pre+1, pre+1+2*n, cmpx);
        printf("%.3f\n", close_pair(1, 2*n));
    }
    return 0;
}

最近點對 - （判斷是否在同一集合）